Is it possible to either prove or disprove the following?
If m is a positive odd integer such that 2^{m} = 2 (mod. m(m-1)) then m is a prime.
Examples:
2^{7} = 2 (mod. 42 = 7x6). 7 is prime.
2^{43} = 2 (mod. 1806 = 43x42). 43 is prime.
Is it possible to either prove or disprove the following?
If m is a positive odd integer such that 2^{m} = 2 (mod. m(m-1)) then m is a prime.
Examples:
2^{7} = 2 (mod. 42 = 7x6). 7 is prime.
2^{43} = 2 (mod. 1806 = 43x42). 43 is prime.
Is the following proof valid?
Theorem.
If m is a positive odd integer such that 2m ≡ 2 (mod. m(m-1)) then m ≡ 3 (mod. 4) and m is a prime.
Proof.
Let 2m ≡ 2 (mod. m(m-1)) then 2m - 2 =km(m-1) for some positive integer k.
This gives 2m-1 – 1 = km(m-1)/2 and since 2m-1 – 1 is odd, (m-1)/2 is odd implying m ≡ 3 (mod. 4).
Let p be the least valued prime dividing (m-1)/2 . Then p is odd and 2m-1 ≡ 1 (mod. p).
Let d be the order of 2 (mod. p).
Then d is a positive integer < p and 2d ≡ 1 (mod. p).
So d divides (m-1) and d divides (p-1).
This implies there is a prime dividing d, less than p, which divides (m-1)/2 .
This contrdicts p is the least valued prime dividing (m-1)/2 .
Hence, m is a prime.